Question

What will be the oxidation state of Carbon in baking soda i.e. $NaHC{O}_3$

• A. $+2$
• B. $+3$
• C. $+4$
• D. $-2$

Answer 1

The correct option is C $+4$

We know that bicarbonate exists as $HC{O}_3^{-}$ ion.
Charge on $Na=+1$
Charge on $H=+1$
Charge on $O=-2$
Overall charge on $NaHC{O}_3=0$
Let oxidation state of $C$ be $x$
$+1+1+x+(3\times (-2\left)\right)=0x=+4$
Theory :
Here are some important radicals with their formulas.
$\text{Radicals}\text{Formula}$
$\text{Nitrate}N{O}_3^{-}$
$\text{Nitrite}N{O}_2^{-}$
$\text{Bisulphate}HS{O}_4^{-}$
$\text{Bicarbonate}HC{O}_3^{-}$
$\text{Cyanide}C{N}^{-}$
$\text{Sulphate}S{O}_4^{2-}$
$\text{Sulphide}{S}^{2-}$
$\text{Bromide}B{r}^{-}$
$\text{Chloride}C{l}^{-}$
$\text{Iodide}{I}^{-}$
$\text{Carbon monoxide}CO$
$\text{Nitric oxide}NO$